Đáp án:
Bài2 :
$a) 2x.(x-9) +3(x-9)=0$
$⇔(x-9)(2x+3)=0$
$⇔$\(\left[ \begin{array}{l}x-9=0\\2x+3=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-9\\x=-\dfrac{3}{2}\end{array} \right.\)
Vậy x ∈ { $-9 ; -\dfrac{3}{2}$}
$b) 2x^3 +3x^2+2x+3=0$
$ = x^2(2x+3) +(2x+3)=0$
$=(2x+3)(x^2+1)=0$
Vì $x^2 ≥0 Nên x^2 +1 >0$
$⇔2x+3 =0$
$⇔2x=-3$
$⇔x=-\dfrac{3}{2}$
Vậy $x=-\dfrac{3}{2}$
$c)x^3 +27 +(x+3)(x-9)=0$
$⇔x^3 +27 +x^2 -9x+3x-27=0$
$⇔x^3 +x^2 -6x=0$
$⇔x^3 -3x^2+2x^2-6x=0$
$⇔x^2(x-3)+2x(x-3)=0$
$⇔(x-3)(x^2+2x)=0$
$⇔x(x+2)(x-3)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x+2=0\\x-3=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=0\\x=-2\\x=3\end{array} \right.\)
Vậy x ∈ { $0 ; -2 ; 3$}
